Question

me ajudem, quero saber se ta certo
considerando x = 0,1666666.., y = 0,25 e z = -1,666666... calcule o valor
x + y + z = 1/99
x - y + z = 31/99
x - (y + z) = -31/99
y + (x- z) = 1/99

Answer 1

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$\sf x=0,1666....\cdot10\\\sf 10x=1,666...\cdot10\\\sf 100x=16,666...\\-\underline{\begin{cases}\sf100x=16,\diagup\!\!\!6\diagup\!\!\!6\diagup\!\!\!6...\\\sf10x=1,\diagup\!\!\!6\diagup\!\!\!6\diagup\!\!\!6\end{cases}}\\\sf 90x=15\\\sf x=\dfrac{15\div15}{90\div15}=\dfrac{1}{6}$

$\sf y=0,25=\dfrac{25\div25}{100\div25}=\dfrac{1}{4}$

$\sf z=1,666...\cdot10\\\sf 10z=16,666...\\-\underline{\begin{cases}\sf10z=16,666...\\\sf z=1,666...\end{cases}}\\\sf 9z=15\\\sf z=\dfrac{15\div3}{9\div3}=\dfrac{5}{3}\\\sf z=-1,666...=-\dfrac{5}{3}$

$\sf x+y+z=\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{5}{3}\\\sf x+y+z=\dfrac{2+3-20}{12}\\\sf x+y+z=-\dfrac{15\div3}{12\div3}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x+y+z=-\dfrac{5}{4}}}}}\\\sf portanto~letra~A~\acute e~falsa.$

$\sf x-y+z=\dfrac{1}{6}-\dfrac{1}{4}-\dfrac{5}{3}\\\sf x-y+z=\dfrac{2-3+20}{12}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x-y+z=\dfrac{19}{12}}}}}\\\sf portanto~alternativa~b~\acute e~falsa.$

$\sf x-(y+z)=\dfrac{1}{6}-\bigg(\dfrac{1}{4}-\dfrac{5}{3}\bigg)\\\sf x-(y+z)=\dfrac{1}{6}-\dfrac{1}{4}+\dfrac{5}{3}\\\sf x-(y+z)=\dfrac{2-3+20}{12}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x-(y+z)=\dfrac{19}{12}}}}}\\\sf alternativa~c~\acute e~falsa.$

$\sf y+(x-z)=\dfrac{1}{4}+\bigg(\dfrac{1}{6}-\bigg[-\dfrac{5}{3}\bigg]\bigg)\\\sf y+(x-z)=\dfrac{1}{4}+\bigg(\dfrac{1}{6}+\dfrac{5}{3}\bigg)\\\sf y+(x-z)=\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{5}{3}\\\sf y+(x-z)=\dfrac{3+2+20}{12}\\\huge\boxed{\boxed{\boxed{\boxed{\sf y+(x-z)=\dfrac{25}{12}}}}}\\\sf alternativa~d~\acute e~falsa.$